Why does TD Learning require Markovian domains?
The Markov assumption is used when deriving the Bellman equation for state values:
$$v(s) = \sum_a \pi(a|s)\sum_{r,s'} p(r,s'|s,a)(r + \gamma v(s'))$$
One requirement for this equation to hold is that $p(r,s'|s,a)$ is consistent. The current state $s$ is a key argument of that function. There is no adjustment for history of previous states, actions or rewards. This is the same as requiring the Markov trait for state, i.e. that $s$ holds all information necessary to predict outcome probabilities of the next step.
The one step TD target that is sampled in basic TD learning is simply the inner part of this:
$$G_{t:t+1} = R_{t+1} + \gamma \hat{v}(S_{t+1})$$
which when sampled is equal to $v(s)$ in expectation *, when $S_t = s$. That is, when you measure a single instance of the TD target and use it to update a value function, you implicitly assume that the values or $r_{t+1}$ and $s_{t+1}$ that you observed occur with probabilities determined by $\pi(a|s)$ and $p(r,s'|s,a)$ as shown by the Bellman equation.
So the theory behind TD learning uses the Markov assumption, otherwise the sampled TD targets would be incorrect.
In practice you can get away with slightly non-Markov environments - most measurements of state for machinery are approximations that ignore details at some level, for instance, and TD learning can solve optimal control in many robotics environments. However, Monte Carlo methods are more robust against state representations that are not fully Markov.

* Technically this sample is biased because $\hat{v}(S_{t+1})$ is not correct when learning starts. The bias reduces over time and multiple updates. So the expected value during learning is approximately the same as the true value as shown by the Bellman equation.